# Can we determine if a complex number is greater than another? [duplicate]

Is it possible to determine if one complex number is greater than another? Or as the question implies is there an "order" to complex numbers (like 1 is before 2 in the real numbers)?

I thought that would could simply use the modulus to determine if one complex numbers is greater than another, though I believe this can't be the only way used (what if 2 complex numbers have the same modulus and are quite different). So I thought, if the point is in the uppermost right quadrant of the complex plane, then both real and imaginary parts are positive, so it would be greater than any other complex number in a different quadrant of the complex plane (you might say what about the modulus, but in the reals $1>-2$ even though $|-2|>|1|$). But what if one complex number has a positive real, and negative imaginary and another one has a negative real and positive imaginary? (And for arguments sake they both have the same modulus)

If we can't determine why not? In the real numbers it seems (to me), quit trivial at a basic level to determine if one real is greater than another e.g. $2>1$. What is this property of numbers called? Why doesn't complex numbers exhibit this property (if indeed it doesn't)?

## marked as duplicate by Hans Lundmark, user91500, Watson, Workaholic, Daniel W. FarlowJun 4 '16 at 12:41

• I'm sure you already know this, but the set of all numbers with the same modulus will trace out a circle in the complex plane. For instance, all numbers with modulus $1$ can be expressed in the form $\exp(i\theta)$ for $-\pi<\theta\le\pi$. – J. M. is a poor mathematician Jun 4 '16 at 1:17
• You can determine if a complex number is greater than another if you first define what you mean by "greater than". – immibis Jun 4 '16 at 10:26

It is possible to order the complex numbers. For instance, one could define $x_1+iy_1<x_2+iy_2$ if $x_1<x_2$ or if $x_1=x_2$ and $y_1<y_2$.

However, it's impossible to define a total order on the complex numbers in such a way that it becomes an ordered field. This is because in an ordered field the square of any non-zero number is $>0$. Hence we would have $-1=i^2>0$, and adding $1$ to both sides would imply $0>1=1^2$, which is a contradiction.

• Ok, why for it to be ordered does the square of any non-zero number have to be greater than 0? I can see how that would make it much easier to order, but why can't it be ordered if it doesn't have that property? – frog1944 Jun 4 '16 at 0:43
• It's a consequence of the axioms for an ordered field. The key is that in an ordered field, the ordering must be compatible with addition and multiplication. Note that the ordering I mentioned in my post isn't compatible with multiplication. – carmichael561 Jun 4 '16 at 0:47
• I think this answer could be improved by stating the axioms that imply $i^2>0$. It's hard to see why "the square of any non-zero number is $>0$" is a reasonable thing to require for an order, but it's easier to agree that it is natural to require "if $a$ and $b$ are positive, then $ab$ is positive" and "if $a<b$ then $a+c<b+c$". – JiK Jun 4 '16 at 12:38
• "It is possible to order the complex numbers. For instance, one could define $x_1+iy_1<x_2+iy_2$ if $x_1<x_2$ or if $x_1=x_2$ and $y_1<y_2$." Is there any particular benefit why one want to define "order" that way? When looking up through my complex analysis notes (notably Lesson 1), we were not even allowed to do just that. – imranfat May 8 '18 at 21:45
• @imranfat: This is called the dictionary or lexicographical order, and comes up in some counterexamples in topology. – carmichael561 May 9 '18 at 5:07

There is no total order on $\mathbb{C}$ compatible with the order on $\mathbb{R}$ and compatible with the algebraic operations. Suppose there was such an order, then either $i>0$ or $i<0$. If $i>0$, then multiplying by $i$ we get that $-1=i^2>0$ which is impossible. If $i<0$, then multiplying by $i$ reverses the inequality, and so we get that $-1=i^2>0$. Both lead to contradictions.

• With regards to the multiplication of $i$, maybe you can't simply multiply it like a real number, so that if I multiply both sides by $i$ it doesn't behave the same way as a real number would. In as much as if I square both sides of an inequality then the inequality flips. So if I have $-1<0$ and I then square both sides I have to flip the inequality, $1>0$. So could it be seen that instead of multiplying both sides by $i$, you're really squaring both sides? – frog1944 Jun 4 '16 at 0:48
• @frog1944: "maybe you can't simply multiply it" -- well then it's not an ordered field. We've established that you can order the complex numbers, this answer is proving that you can't order them compatibly with algebraic operations. If you don't need the order to be compatible with multiplication then use carmichael's: it plays nicely with addition but not multiplication. – Steve Jessop Jun 4 '16 at 1:16
• Ok, thanks @SteveJessop – frog1944 Jun 4 '16 at 1:21
• What you've been exploring is quite a significant result, though, which is that the real numbers is the only ordered field that satisfies the so-called Archimedean property (for any element there is a larger integer, that is no infinite values) and also is complete (meaning every Cauchy sequence converges). So one answer to "Why doesn't complex numbers exhibit this property" is "because they form a complete field that isn't isomorphic to the reals". – Steve Jessop Jun 4 '16 at 1:31

The field $\mathbb{C}$ has only a total order compatible with addition. It has no total order which is compatible with multiplication, prohibiting it from being an ordered field. Finally, its usual topology cannot be generated by any total order.

If $i>0$, then $i^2>0 \Rightarrow -1>0$ which is a contradiction.
If $i=0$, then $i^2=0 \Rightarrow -1=0$ which is a contradiction.
If $i<0$, then $i^2>0 \Rightarrow -1>0$ which is a contradiction.

Thus $ib$ is neither greater nor equal nor less than $0$.

So any complex number $a+ib$, ($a,b \in \mathbb{R}$) is neither greater nor equal nor less than $a$.
And the reasoning follows.